## Bar Bending Schedule of One way Slab

Here we calculate bar bending schedule of one way slab given in detailed explanation:

Main Reinforcement bars we provide in the short span of the slabs and the distribution bars will along the longer span of the slab.

For your better understanding,let’s take this example.

Suppose we have a one way slab,which having a length 5 m or width 2 m (clear span).The main bars will be 12 mm in diameter @100 mm c/c spacing.The Distribution bars will be 8 mm in diameter @125mm c/c spacing.The clear cover will be 25 mm(Top or Bottom) and the thickness of the slab is 150 mm.

1.Calculate the quantity of steel?

2.Calculate the weight of steel?

### Given data:

Length = 5 m (5000 mm)

Width = 2 m (2000 mm)

Main Bar = 12 mm @ 100 mm c/c

Distribution Bar = 8 mm @ 125 mm c/c

Clear cover = 25 mm from (Top and Bottom)

**Calculation of Brick Work Quantity**

### Solution:

The quantity is done into two steps.

1.Calculation of number of bars

2.Cutting length of bars

Step 1:Calculation of bars no’s

First calculate the number of bars required (main and distribution both).

### Formula:

= (Total length – clear cover)/center to center spacing + 1

### Main bar:

= (5000 – (25+25))/100 + 1

= 4950 / 100 + 1

= 51 bars

### Distribution bar:

= (2000 – ( 25 + 25))/125 + 1

= 1950/125+1

= 17 bars

### Step 2.Cutting Length

#### Main Bar:

#### Formula

= (L) + (2 x Ld) + (1 x 0.42D) – (2 x 1d)

#### Where

L = Clear span of the slab

Ld = Development Length which is 40d ( where d is diameter of bar)

0.42D = Inclined length (Bend length)

1d = 45 degree bends (d is diameter of bar)

**Bar Bending Schedule of Column**

#### First calculate the length of “D”

= (Thickness) – 2 (clear cover at Top,Bottom) – Diameter of the bar

= 150 – 2(25) – 12

= 88 mm

By putting values

### Cutting Length:

= 2000 + (2 x 40 x 12) + (1 x 0.42 x 88) – (2 x 1 x 12)

= 2000 + 960 + 36.96 – 24

= 2972.96 mm or 2.973 m

### Distribution Bar:

= Clear span + ( 2 x Development length (Ld))

= 5000 + (2 x 40 x 8)

= 5640 mm or 5.64 m

### Conclusion:

### 1.Main Bar:

= 51 nos

#### Length:

= 51 x 2.973 m

= 151.623 m

#### Weight:

= (D2/162) x length

= 134.776 kg

### 2.Distribution Bar:

= 17 nos

Note: The weight of the bar may vary depending upon the properties of the steel.

**Bar Bending Schedule of Column Footing**